Equation Of A Circle Examples With Answers

The Equation of a Circle: Examples and Solutions

Circles are fundamental geometric shapes with a rich history and numerous applications in mathematics, science, and engineering. Understanding the equation of a circle is essential for solving a wide range of problems, from finding the distance between two points to designing circular structures.

In this article, we will delve into the equation of a circle, exploring its various forms, properties, and applications. We will also provide numerous examples with detailed solutions to help you master this important concept.

Comprehensive Overview

A circle is defined as the set of all points in a plane that are equidistant from a fixed point called the center. The distance from the center to any point on the circle is called the radius.

The equation of a circle is a mathematical expression that describes the relationship between the coordinates of any point on the circle and the center and radius of the circle.

Standard Form of the Equation of a Circle

The standard form of the equation of a circle with center (h, k) and radius r is:

(x - h)^2 + (y - k)^2 = r^2

This equation is derived from the distance formula, which states that the distance between two points (x1, y1) and (x2, y2) is:

√((x2 - x1)^2 + (y2 - y1)^2)

In the case of a circle, the distance between any point (x, y) on the circle and the center (h, k) is equal to the radius r. Therefore, we have:

√((x - h)^2 + (y - k)^2) = r

Squaring both sides of the equation, we obtain the standard form of the equation of a circle:

(x - h)^2 + (y - k)^2 = r^2

General Form of the Equation of a Circle

The general form of the equation of a circle is:

x^2 + y^2 + 2gx + 2fy + c = 0

where g, f, and c are constants.

To convert the general form of the equation of a circle to the standard form, we need to complete the square for both the x and y terms.

Completing the square involves adding and subtracting the square of half the coefficient of the x term and the y term to the equation.

For example, to complete the square for the x term in the general form of the equation of a circle, we add and subtract (g^2) to the equation:

x^2 + 2gx + g^2 - g^2 + y^2 + 2fy + c = 0

We can then rewrite the first three terms as a perfect square:

(x + g)^2 - g^2 + y^2 + 2fy + c = 0

Similarly, to complete the square for the y term, we add and subtract (f^2) to the equation:

(x + g)^2 - g^2 + y^2 + 2fy + f^2 - f^2 + c = 0

We can then rewrite the first three terms as a perfect square:

(x + g)^2 - g^2 + (y + f)^2 - f^2 + c = 0

Rearranging the terms, we obtain the standard form of the equation of a circle:

(x + g)^2 + (y + f)^2 = g^2 + f^2 - c

Therefore, the center of the circle is (-g, -f) and the radius is √(g^2 + f^2 - c).

Examples with Answers

Here are some examples of finding the equation of a circle:

Example 1: Find the equation of the circle with center (2, 3) and radius 5.

Solution:

Using the standard form of the equation of a circle, we have:

(x - 2)^2 + (y - 3)^2 = 5^2

Simplifying, we get:

(x - 2)^2 + (y - 3)^2 = 25

Example 2: Find the equation of the circle with center (-1, 4) and radius 2.

Solution:

Using the standard form of the equation of a circle, we have:

(x + 1)^2 + (y - 4)^2 = 2^2

Simplifying, we get:

(x + 1)^2 + (y - 4)^2 = 4

Example 3: Find the equation of the circle with center (0, 0) and radius 7.

Solution:

Using the standard form of the equation of a circle, we have:

(x - 0)^2 + (y - 0)^2 = 7^2

Simplifying, we get:

x^2 + y^2 = 49

Example 4: Find the equation of the circle with center (5, -2) and radius 3.

Solution:

Using the standard form of the equation of a circle, we have:

(x - 5)^2 + (y + 2)^2 = 3^2

Simplifying, we get:

(x - 5)^2 + (y + 2)^2 = 9

Example 5: Find the equation of the circle with center (-3, -1) and radius 4.

Solution:

Using the standard form of the equation of a circle, we have:

(x + 3)^2 + (y + 1)^2 = 4^2

Simplifying, we get:

(x + 3)^2 + (y + 1)^2 = 16

Here are some examples of finding the center and radius of a circle from its equation:

Example 6: Find the center and radius of the circle with equation (x - 1)^2 + (y + 2)^2 = 9.

Solution:

Comparing the equation to the standard form, we have:

(x - h)^2 + (y - k)^2 = r^2

Therefore, the center of the circle is (1, -2) and the radius is 3.

Example 7: Find the center and radius of the circle with equation (x + 3)^2 + (y - 4)^2 = 16.

Solution:

Comparing the equation to the standard form, we have:

(x - h)^2 + (y - k)^2 = r^2

Therefore, the center of the circle is (-3, 4) and the radius is 4.

Example 8: Find the center and radius of the circle with equation x^2 + y^2 = 25.

Solution:

Comparing the equation to the standard form, we have:

(x - h)^2 + (y - k)^2 = r^2

Therefore, the center of the circle is (0, 0) and the radius is 5.

Example 9: Find the center and radius of the circle with equation x^2 + y^2 - 4x + 6y - 12 = 0.

Solution:

To find the center and radius of the circle, we need to convert the equation to the standard form. To do this, we complete the square for both the x and y terms.

Completing the square for the x term, we add and subtract (4/2)^2 = 4 to the equation:

x^2 - 4x + 4 - 4 + y^2 + 6y - 12 = 0

We can then rewrite the first three terms as a perfect square:

(x - 2)^2 - 4 + y^2 + 6y - 12 = 0

Completing the square for the y term, we add and subtract (6/2)^2 = 9 to the equation:

(x - 2)^2 - 4 + y^2 + 6y + 9 - 9 - 12 = 0

We can then rewrite the first three terms as a perfect square:

(x - 2)^2 - 4 + (y + 3)^2 - 9 - 12 = 0

Rearranging the terms, we obtain the standard form of the equation of a circle:

(x - 2)^2 + (y + 3)^2 = 25

Therefore, the center of the circle is (2, -3) and the radius is 5.

Example 10: Find the center and radius of the circle with equation 2x^2 + 2y^2 + 8x - 12y + 10 = 0.

Solution:

First, divide the entire equation by 2 to simplify:

x^2 + y^2 + 4x - 6y + 5 = 0

Now, complete the square for both x and y terms:

(x^2 + 4x) + (y^2 - 6y) = -5

To complete the square for x, add and subtract (4/2)^2 = 4:

(x^2 + 4x + 4) - 4

To complete the square for y, add and subtract (-6/2)^2 = 9:

(y^2 - 6y + 9) - 9

Rewrite the equation:

(x^2 + 4x + 4) + (y^2 - 6y + 9) - 4 - 9 = -5

Now, rewrite as perfect squares:

(x + 2)^2 + (y - 3)^2 = -5 + 4 + 9

Simplify:

(x + 2)^2 + (y - 3)^2 = 8

Therefore, the center of the circle is (-2, 3) and the radius is √8 = 2√2.

Example 11: Determine the equation of a circle if the endpoints of its diameter are A(2,5) and B(6,-3).

Solution:

The center of the circle is the midpoint of the diameter. The midpoint M(x, y) of two points A(x1, y1) and B(x2, y2) is given by: x = (x1 + x2)/2 and y = (y1 + y2)/2

In this case: x = (2 + 6)/2 = 4 y = (5 + (-3))/2 = 1

So, the center of the circle is (4, 1).

The radius of the circle is half the length of the diameter. We can find the length of the diameter (distance between A and B) using the distance formula: Distance = √((x2 - x1)^2 + (y2 - y1)^2) Distance = √((6 - 2)^2 + (-3 - 5)^2) Distance = √(4^2 + (-8)^2) Distance = √(16 + 64) Distance = √80 = 4√5

The radius is half the distance, so r = (4√5)/2 = 2√5

Now, using the standard equation of a circle: (x - h)^2 + (y - k)^2 = r^2 (x - 4)^2 + (y - 1)^2 = (2√5)^2 (x - 4)^2 + (y - 1)^2 = 20

Therefore, the equation of the circle is (x - 4)^2 + (y - 1)^2 = 20.

Example 12: A circle has the equation (x - 3)^2 + (y + 2)^2 = 25. Determine if the point (6, 1) lies on the circle, inside the circle, or outside the circle.

Solution:

To determine the position of the point (6, 1) with respect to the circle, substitute the coordinates of the point into the equation of the circle and compare the result to the radius squared.

(x - 3)^2 + (y + 2)^2 = 25 (6 - 3)^2 + (1 + 2)^2 = ? (3)^2 + (3)^2 = ? 9 + 9 = 18

Since 18 < 25, the point (6, 1) lies inside the circle.

Example 13: Find the equation of the circle passing through the points (1, 1), (2, -1), and (3, 2).

Solution:

Let the equation of the circle be x^2 + y^2 + 2gx + 2fy + c = 0. Since the circle passes through the given points, these points must satisfy the equation. Substituting the points one by one we get:

For (1, 1): 1 + 1 + 2g + 2f + c = 0 => 2g + 2f + c = -2 (Equation 1)

For (2, -1): 4 + 1 + 4g - 2f + c = 0 => 4g - 2f + c = -5 (Equation 2)

For (3, 2): 9 + 4 + 6g + 4f + c = 0 => 6g + 4f + c = -13 (Equation 3)

Now, solve the system of equations.

Subtract Equation 1 from Equation 2: (4g - 2f + c) - (2g + 2f + c) = -5 - (-2) 2g - 4f = -3 (Equation 4)

Subtract Equation 1 from Equation 3: (6g + 4f + c) - (2g + 2f + c) = -13 - (-2) 4g + 2f = -11 (Equation 5)

Multiply Equation 4 by 1 and Equation 5 by 2:

2g - 4f = -3 8g + 4f = -22

Add the two equations: 10g = -25 g = -2.5

Substitute g = -2.5 into Equation 4: 2(-2.5) - 4f = -3 -5 - 4f = -3 -4f = 2 f = -0.5

Substitute g = -2.5 and f = -0.5 into Equation 1: 2(-2.5) + 2(-0.5) + c = -2 -5 - 1 + c = -2 c = 4

So the equation of the circle is: x^2 + y^2 - 5x - y + 4 = 0.

Tips & Expert Advice

• Always remember the standard form of the equation of a circle: (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius.
• To find the equation of a circle, you need to know the center and radius.
• If you are given the general form of the equation of a circle, you need to complete the square to convert it to the standard form.
• When working with circles, always draw a diagram to help you visualize the problem.

FAQ (Frequently Asked Questions)

• What is the equation of a circle?

  The equation of a circle is a mathematical expression that describes the relationship between the coordinates of any point on the circle and the center and radius of the circle.

• What is the standard form of the equation of a circle?

  The standard form of the equation of a circle with center (h, k) and radius r is:

  (x - h)^2 + (y - k)^2 = r^2
  

• What is the general form of the equation of a circle?

  The general form of the equation of a circle is:

  x^2 + y^2 + 2gx + 2fy + c = 0
  

• How do I find the center and radius of a circle from its equation?

  To find the center and radius of a circle from its equation, you need to convert the equation to the standard form. This can be done by completing the square for both the x and y terms.

• What are some applications of circles?

  Circles have numerous applications in mathematics, science, and engineering, including:

  • Finding the distance between two points
  • Designing circular structures
  • Calculating the area and circumference of circular objects
  • Modeling periodic phenomena, such as the motion of a pendulum

Conclusion

Understanding the equation of a circle is essential for solving a wide range of problems in mathematics, science, and engineering. By mastering the concepts and techniques discussed in this article, you will be well-equipped to tackle any problem involving circles. Practice is key, so work through as many examples as possible to solidify your understanding.

How do you feel about the equation of a circle now? Are you ready to apply this knowledge to real-world problems?